Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $z = \dfrac{k^2 - 6k + 9}{k^2 + 2k + 1} \div \dfrac{k^2 - 3k}{k^2 + k} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $z = \dfrac{k^2 - 6k + 9}{k^2 + 2k + 1} \times \dfrac{k^2 + k}{k^2 - 3k} $ First factor out any common factors. $z = \dfrac{k^2 - 6k + 9}{k^2 + 2k + 1} \times \dfrac{k(k + 1)}{k(k - 3)} $ Then factor the quadratic expressions. $z = \dfrac {(k - 3)(k - 3)} {(k + 1)(k + 1)} \times \dfrac {k(k + 1)} {k(k - 3)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac { (k - 3)(k - 3) \times k(k + 1)} { (k + 1)(k + 1) \times k(k - 3)} $ $z = \dfrac {k(k - 3)(k - 3)(k + 1)} {k(k + 1)(k + 1)(k - 3)} $ Notice that $(k + 1)$ and $(k - 3)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac {k(k - 3)(k - 3)\cancel{(k + 1)}} {k\cancel{(k + 1)}(k + 1)(k - 3)} $ We are dividing by $k + 1$ , so $k + 1 \neq 0$ Therefore, $k \neq -1$ $z = \dfrac {k\cancel{(k - 3)}(k - 3)\cancel{(k + 1)}} {k\cancel{(k + 1)}(k + 1)\cancel{(k - 3)}} $ We are dividing by $k - 3$ , so $k - 3 \neq 0$ Therefore, $k \neq 3$ $z = \dfrac {k(k - 3)} {k(k + 1)} $ $ z = \dfrac{k - 3}{k + 1}; k \neq -1; k \neq 3 $